Completely N-continuous Multifunctions
Multifunciones completamente N-continuas Erdal Ekici ([email protected])
Department of Mathematics, Canakkale Onsekiz Mart University,
Terzioglu Campus 17020 Canakkale – TURKEY
Abstract
In this paper, the concept of completely N-continuous multifunction is defined and their basic characterizations and properties are obtained.
Key words and phrases: complete N-continuity, multifunction.
Resumen
En este art´ıculo se define el concepto de multifunci´on completamente N-continua y se obtienen su caracterizaci´on y propiedades b´asicas.
Palabras y frases clave:N-continuidad completa, multifunci´on.
1 Introduction
In recent years, various classes of weak and strong forms of continuity have been major interest among General Topologists. The aim of this paper is to present and study the notion of completely N-continuous multifunctions.
A topological space (X, τ) is called nearly compact [7] if every cover ofX by regular open sets has a finite subcover.
Let (X, τ) be a topological space and let A be a subset of X. If every cover of A by regular open subsets of (X, τ) has a finite subfamily whose union covers A, then A is called N-closed (relative to X) [4]. Sometimes, such sets are called N-sets or α-nearly compact. The class of N-closed sets is important in the study of functions with strongly closed graphs [5].
Received 2003/07/10. Revised 2005/02/18. Accepted 2005/03/18.
MSC (2000): 54C05, 54C60, 54C10.
Basic observation about N-closed sets involve the fact that every compact set is N-closed.
On the other hand, many authors studied N-closed sets and their topolo- gies [2, 3, etc.] in the literature.
In this paper, a multifunctionF :X →Y from a topological spaceX to a topological space Y is a point to set correspondence and is assumed that F(x)6=∅ for allxwhere∅denotes the empty set.
IfAis a subset of a topological space, thencl(A) denote the closure ofA andint(A) denote the interior ofAandco(A) denote the complement ofA.
For a multifunction F : X → Y from a topological space (X, τ) to a topological space (Y, υ), we denote the upper and lower inverse of a subset V of Y byF+(V) andF−(V), respectively;F+(V) ={x∈X :F(x)⊆V}and F−(V) ={x∈X :F(x)∩V 6=∅}[1].
A multifunctionF :X →Y is said to be (i) upper semi continuous at a pointx∈X if for each open setV inY withF(x)⊆V, there exists an open set U containingxsuch that F(U)⊆V; and (ii) lower semi continuous at a point x∈X if for each open setV in Y with F(x)∩V 6=∅, there exists an open set U containingxsuch thatF(u)∩V 6=∅ for everyu∈U [6].
For a multifunctionF :X →Y, the graph multifunctionGF :X →X×Y is defined asGF(x) ={x} ×F(x) for everyx∈X [8].
2 Completely N-continuous multifunctions
Definition 1. Let F : X → Y be a multifunction from a topological space (X, τ)to a topological space(Y, υ). F is said to be
(1) lower completely N-continuous if for each x∈X and for each open set V such that x∈ F−(V), there exists an open set U containing x and having N-closed complement such thatU ⊆F−(V),
(2) upper completely N-continuous if for each x∈X and for each open set V such that x∈ F+(V), there exists an open set U containing x and having N-closed complement such thatU ⊆F+(V).
We know that a net (xα) in a topological space (X, τ) is called eventually in the set U ⊆X if there exists an index α0 ∈ J such that xα ∈ U for all α≥α0.
Definition 2. Let(X, τ)be a topological space and let(xα)be a net inX. It is said that the net (xα)n-converges tox(writtenxα→nx) if for each open
set G ⊆X having N-closed complement, there exists an index α0 ∈ I such that xα∈Gfor each α≥α0.
Definition 3. Let (X, τ)be a topological space. A set Gin X said to be n- open if for eachx∈G,there exists an open setH having N-closed complement such that x∈H andH ⊆G. The complement of a n-open set is called to as a n-closed set.
The following theorem gives us some characterizations of lower (upper) completely N-continuous multifunction.
Theorem 4. Let F : X → Y be a multifunction from a topological space (X, τ)to a topological space (Y, υ). Then the following statements are equiv- alent:
i-) F is lower (upper) completely N-continuous,
ii-) For each x ∈ X and for each open set V such that F(x)∩V 6= ∅ (F(x) ⊆ V), there exists an open set U containing x and having N- closed complement such that ify∈U, then F(y)∩V 6=∅ (F(y)⊆V), iii-) For each x ∈ X and for each closed set K such that x ∈ F−(coK)
(x∈F+(coK)), there exists a closed N-closed setH such thatx∈co(H) andF+(K)⊆H (F−(K)⊆H),
iv-) F−(V) (F+(V)) is a n-open set for any open set V ⊆Y, v-) F+(K)(F−(K)) is a n-closed set for any closed set K⊆Y,
vi-) For each x ∈ X and for each open set V such that F(x)∩V 6= ∅ (F(x)⊆V), there exists a n-open setU containingxsuch that ify∈U, thenF(y)∩V 6=∅ (F(y)⊆V),
vii-) For eachx∈X and for each net(xα)which n-converges to xinX and for each open set V ⊆Y such that x∈ F−(V) (x ∈F+(V)), the net (xα)is eventually inF−(V) (F+(V)).
Proof. (i)⇔(ii): Obvious.
(i)⇔(iii): Let x∈ F−(coK) and letK be a closed set. From (i), there exists an open setU containingxand having N-closed complement such that U ⊆ F−(coK). It follows that co(F−(coK)) = F+(K) ⊆ co(U). We take H =co(U). Thenx∈co(H) andF+(K)⊆H.
The converse can be shown similarly.
(i)⇔(iv): LetV ⊆Y be an open set and letx∈F−(V). From (i), there exists an open setU containingxand having N-closed complement such that U ⊆F−(V). It follows thatF−(V) is a n-open set.
The converse can be obtain similarly from the definition of n-open set.
(iv)⇔(v): Let K be a closed set. Thenco(K) is an open set. From (iv), F−(co(K)) =co(F+(K)) is a n-open set. It follows thatF+(K) is a n-closed set.
The converse can be shown similarly.
(ii)⇔(vi): Let x ∈ X and let V be an open set such that F(x)∩V 6=
∅. From (ii), there exists an open set U containing xand having N-closed complement such that ify ∈U, then F(y)∩V 6=∅. Since U is a n-open set, it follows that (vi) holds.
The converse is similar.
(i)⇒(vii): Let (xα) be a net which n-converges toxinX and letV ⊆Y be any open set such thatx∈F−(V). SinceF is lower completely N-continuous, it follows that there exists an open set U ⊆X containing xand having N- closed complement such that U ⊆ F−(V). Since (xα) n-converges to x, it follows that there exists an index α0 ∈ J such thatxα ∈ U for all α≥α0. From here, we obtain that xα ∈U ⊆F−(V) for allα ≥α0. Thus, the net (xα) is eventually inF−(V).
(vii)⇒(i): Suppose that (i) is not true. There exist a point x and an open setV withx∈F−(V) such that U *F−(V) for each open setU ⊆X containingxand having N-closed complement. LetxU ∈U andxU ∈/F−(V) for each open set U ⊆ X containing x and having N-closed complement.
Then for the neighbourhood net (xU),xU →nx, but (xU) is not eventually in F−(V). This is a contradiction. Thus,F is lower completely N-continuous.
The proof of the upper completely continuity ofF is similar to the above.
Remark 5. For a multifunction F :X →Y from a topological space(X, τ) to a topological space (Y, υ), the following implication hold:
F is lower (upper) completely N-continuous ⇒ F is lower (upper) semi continuous.
However the converse is not true in general by the following example.
Example 6. Take the discrete topologyτDonR. We define the multifunction as the follows; F : (R, τD)→(R, τD), F(x) ={x} for eachx∈X. Then F is lower (upper) semi continuous but F is not lower (upper) completely N- continuous.
Definition 7. Let(X, τ)be a topological space. (X, τ)is called n-regular space if (X, τ)has a base consisting of open sets having N-closed complements.
The following theorem gives us the condition for the converse.
Theorem 8. Let F :X →Y be a multifunction from a n-regular topological space(X, τ)to a topological space(Y, υ). IfFis upper (lower) semi continuous multifunction, then F is upper (lower) completely N-continuous.
Proof. Since each open set is a n-open set in a n-regular space, the proof is obvious.
Definition 9. Let(X, τ)be a topological space and letA⊆X. A pointx∈X is said to be a n-adherent point ofA if each open set containingxand having N-closed complement intersects A. The set of all n-adherent points of A is denoted by n-cl(A).
It is obvious that the setAis n-closed if and only if n-cl(A) =A.
Theorem 10. Let F : X → Y be a multifunction from a topological space (X, τ)to a topological space(Y, υ). F is lower completely N-continuous if and only ifF(n-cl(A))⊂cl(F(A))for each A⊂X.
Proof. Suppose thatF is lower completely N-continuous andA ⊂X. Since cl(F(A)) is a closed set, it follows that F+(cl(F(A))) is a n-closed set in X from Theorem 4. SinceA⊂F+(cl(F(A))), then n-cl(A)⊂n-cl(F+(cl(F(A)))) = F+(cl(F(A))). Thus, we obtain that F(n-cl(A)) ⊂ F(F+(cl(F(A)))) ⊂ cl(F(A)).
Conversely, suppose that F(n-cl(A)) ⊂ cl(F(A)) for each A ⊂ X. Let K be any closed set in Y. Then F(n-cl(F+(K))) ⊂ cl(F(F+(K))) and cl(F(F+(K))) ⊂ cl(K) = K. Hence, n-cl(F+(K)) ⊂ F+(K) which shows that F is lower completely N-continuous.
Theorem 11. Let F : X → Y be a multifunction from a topological space (X, τ)to a topological space(Y, υ). F is lower completely N-continuous if and only if n-cl(F+(B))⊂F+(cl(B))for eachB⊂Y.
Proof. Suppose that F is lower completely N-continuous and B ⊂Y. Then F+(cl(B)) is n-closed in X and F+(cl(B)) =n-cl(F+(cl(B))). Hence, n- cl(F+(B))⊂F+(cl(B)).
Conversely, letKbe any closed set inY. Then n-cl(F+(K))⊂F+(cl(K)) = F+(K)⊂n-cl(F+(K)). Thus, F+(K) =n-cl(F+(K)) which shows thatF is lower completely N-continuous.
Theorem 12. Let F : X → Y be a multifunction from a topological space (X, τ) to a topological space (Y, υ) and let F(X) be endowed with subspace topology. If F is upper completely N-continuous, then F : X → F(X) is upper completely N-continuous.
Proof. SinceF is upper completely N-continuous,F+(V∩F(X)) =F+(V)∩ F+(F(X)) =F+(V) is n-open for each open subsetV ofY. Hence F:X → F(X) is upper completely N-continuous.
Suppose that (X, τ), (Y, υ) and (Z, ω) are topological spaces. It is known that ifF1:X→Y andF2:Y →Zare multifunctions, then the multifunction F2◦F1:X→Z is defined by (F2◦F1)(x) =F2(F1(x)) for eachx∈X. Theorem 13. Let(X, τ),(Y, υ),(Z, ω)be topological spaces and letF :X → Y andG:Y →Zbe multifunctions. IfF :X →Y is upper (lower) completely N-continuous and G:Y →Z is upper (lower) semi continuous, thenG◦F : X →Z is an upper (lower) completely N-continuous multifunction.
Proof. Let V ⊆ Z be any open set. From the definition of G◦F, we have (G◦F)+(V) =F+(G+(V)) ((G◦F)−(V) =F−(G−(V))). SinceGis upper (lower) semi continuous multifunction, it follows that G+(V) (G−(V)) is an open set. Since F is upper (lower) completely N-continuous multifunction, it follows that F+(G+(V)) (F−(G−(V))) is a n-open set. It shows thatG◦F is an upper (lower) completely N-continuous multifunction.
Corollary 14. Let(X, τ),(Y, υ),(Z, ω)be topological spaces and letF :X → Y andG:Y →Zbe multifunctions. IfF :X →Y is upper (lower) completely N-continuous and G:Y →Z is upper (lower) completely N-continuous, then G◦F :X →Z is an upper (lower) completely N-continuous multifunction.
Theorem 15. Let F : X → Y be a multifunction from a topological space (X, τ)to a topological space(Y, υ). Then the graph multifunction ofF is upper completely N-continuous if and only ifF is upper completely N-continuous and X is n-regular space.
Proof. (⇒): Suppose thatGF is upper completely N-continuous. From The- orem 13, F = PY ◦GF is upper completely N-continuous where PY is the projectionX×Y ontoY.
Let U be any open set in X and let U ×Y be an open set containing GF(x). Since GF upper completely N-continuous, there exists an open set V containing x and having N-closed complement such that if x ∈ V, then GF(x)⊂U×Y. Thus, x∈V ⊂U which shows thatU is a n-open set and X is a n-regular space.
(⇐): Letx∈X and letW be an open set containingGF(x). There exist open sets U ⊂X andV ⊂Y such that (x, F(x))⊂U ×V ⊂ W. Since X is n-regular space, there exists an open setG1⊆X containingxand having N-closed complement such thatx∈G1⊂U. SinceF is upper completely N- continuous, there exists an open setG2inXcontainingxand having N-closed complement such that if a∈G2, thenF(a)⊂V. LetH =G1∩G2. ThenH is an open set containing x and having N-closed complement andGF(H)⊂ U ×V ⊂W which implies thatGF is upper completely N-continuous.
Theorem 16. Suppose that(X, τ)and(Xα, τα)are topological spaces where α ∈ J. Let F : X → Q
α∈J
Xα be a multifunction from X to the product space Q
α∈J
Xα and let Pα : Q
α∈J
Xα → Xα be the projection for each α ∈ J. If F is upper (lower) completely N-continuous, then Pα◦F is (lower) upper completely N-continuous for each α∈J.
Proof. Take any α0 ∈ J. Let Vα0 be an open set in (Xα0, τα0). Then (Pα0◦F)+(Vα0) =F+(Pα+0(Vα0)) =F+(Vα0× Q
α6=α0
Xα) (respectively, (Pα0◦ F)−(Vα0) =F−(Pα−0(Vα0)) =F−(Vα0× Q
α6=α0
Xα)). SinceF is upper (lower) completely N-continuous multifunction and sinceVα0× Q
α6=α0
Xαis an open set, it follows thatF+(Vα0× Q
α6=α0
Xα) (respectively,F−(Vα0× Q
α6=α0
Xα)) is n-open in (X, τ). It shows thatPα0◦F is upper (lower) completely N-continuous.
Hence, we obtain that Pα◦F is upper (lower) completely N-continuous for eachα∈J.
Let (X, τ) be a topological space. It is known that the collection of all open subsets having N-closed complements of (X, τ) is a base for a topology τ∗ onX.
Theorem 17. Let F : X → Y be a multifunction from a topological space (X, τ)to a topological space(Y, υ). ThenF : (X, τ)→(Y, υ)is upper (lower) completely N-continuous if and only if F : (X, τ∗)→(Y, υ) is upper (lower) semi continuous.
Proof. Let V ⊆Y be an open set and let x∈ F+(V). Since F : (X, τ) → (Y, υ) is upper completely N-continuous, it follows that there exists an open set U containingxand having N-closed complement such thatU ⊆F+(V).
From here,U ∈τ∗. Thus,F : (X, τ∗)→(Y, υ) is upper semi continuous.
Converse is similar.
The proof of the lower continuity ofF is similar to the above.
Theorem 18. Let(X, τ)be a topological space. Then the following statements are equivalent:
i-)(X, τ)is a n-regular space,
ii-) Each upper (lower) semi continuous multifunction from(X, τ) into a topological space (Y, υ)is upper (lower) completely N-continuous.
Proof. (i)⇒(ii): LetF :X →Y be an upper semi continuous multifunction.
Let V ⊆ Y be an open set and let x∈ F+(V). Then there exists an open set U containingxsuch thatU ⊆F+(V). Since (X, τ) is a n-regular space, it follows that there exists an open set G containingxand having N-closed complement such that G⊆U ⊆F+(V). Thus, we obtain tahat F is upper completely N-continuous.
(ii)⇒(i): Take (Y, υ) = (X, τ). Then the identity multifunctionIXonX is upper semi continuous and henceIX is upper completely N-continuous. From Theorem 17, IX : (X, τ∗)→ (X, τ) is upper semi continuous. Take U ∈ τ.
Then IX+(U) =U ∈τ∗ and it follows thatτ ⊂τ∗. Therefore, τ =τ∗ and we obtain that (X, τ) is a n-regular space.
The proof interested in the lower continuity ofFis similar to the above.
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